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Sequences and Series question

2020 · 6 Sep · Shift 2 · Q31
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Sequences and Series question

2020 · 6 Sep · Shift 2 · Q31

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
The common difference of the A.P. b1, b2, … , bm is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
  1. A
    127
  2. B
    81
  3. C
    –127
  4. D
    -81
View written solutionFree

Correct answer: D

  1. Let the A.P. a1,a2,…,ana_1,a_2,\dots,a_na1​,a2​,…,an​ have first term aaa and common difference ddd.

    Then ak=a+(k−1)da_k=a+(k-1)dak​=a+(k−1)d

  2. Given: a40=−159,a100=−399a_{40}=-159,\qquad a_{100}=-399a40​=−159,a100​=−399

    So, a+39d=−159a+39d=-159a+39d=−159 a+99d=−399a+99d=-399a+99d=−399

  3. Subtract the first equation from the second: 60d=−24060d=-24060d=−240 d=−4d=-4d=−4

  4. Now find aaa using a+39d=−159a+39d=-159a+39d=−159: a+39(−4)=−159a+39(-4)=-159a+39(−4)=−159 a−156=−159a-156=-159a−156=−159 a=−3a=-3a=−3

    Hence, ak=−3+(k−1)(−4)=1−4ka_k=-3+(k-1)(-4)=1-4kak​=−3+(k−1)(−4)=1−4k

  5. Compute a70a_{70}a70​: a70=−3+69(−4)=−3−276=−279a_{70}=-3+69(-4)=-3-276=-279a70​=−3+69(−4)=−3−276=−279

    Given: b100=a70=−279b_{100}=a_{70}=-279b100​=a70​=−279

  6. Let the A.P. b1,b2,…,bmb_1,b_2,\dots,b_mb1​,b2​,…,bm​ have first term bbb and common difference dbd_bdb​.

    The common difference of the bbb-A.P. is 2 more than that of the aaa-A.P., so db=d+2=−4+2=−2d_b=d+2=-4+2=-2db​=d+2=−4+2=−2

  7. Using b100=b+99dbb_{100}=b+99d_bb100​=b+99db​: b+99(−2)=−279b+99(-2)=-279b+99(−2)=−279 b−198=−279b-198=-279b−198=−279 b=−81b=-81b=−81

  8. Therefore, b1=−81b_1=-81b1​=−81

  9. Checking options:

    • A: 127127127 ✗
    • B: 818181 ✗
    • C: −127-127−127 ✗
    • D: −81-81−81 ✓

So the correct answer is Option D.

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